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From: Tim Rentsch <tr.17687@z991.linuxsc.com>
Newsgroups: comp.lang.c
Subject: Re: question about nullptr
Date: Mon, 12 Aug 2024 17:42:39 -0700
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James Kuyper <jameskuyper@alumni.caltech.edu> writes:

> On 7/10/24 17:23, Keith Thompson wrote:
>
>> Tim Rentsch <tr.17687@z991.linuxsc.com> writes:
>>
>>> Keith Thompson <Keith.S.Thompson+u@gmail.com> writes:
>>>
>>>> Tim Rentsch <tr.17687@z991.linuxsc.com> writes:
>>>> [...]
>>>>
>>>>> This posting has inspired me to try using (long)0.0
>>>>> whenever a null pointer constant is needed.  As for
>>>>> example
>>>>>
>>>>> (void*){ (long)0.0 }
>>>>>
>>>>> as an argument to a variadic function where a pointer
>>>>> is expected.
>>>>
>>>> But surely ((void*)('/'/'/'-'/'/'/')) is more elegant.
>>>
>>> Surely not.  Furthermore the form I showed has a point,
>>> whereas this example is roughly the equivalent of a
>>> first grade knock-knock joke.
>>
>> I was of course joking.  I assumed you were as well.
>>
>> What is the point of (void*){ (long)0.0 }?  I don't believe it's
>> a null pointer constant even in C23.
>
> I think you're right about that.

The compound literal is not a null pointer constant.  I didn't
say it is.  The null pointer constant is (long)0.0, like my
earlier posting pointed out.

> "An integer constant expression132) ... shall only have operands
> that are ... compound literal constants of arithmetic type that
> are the immediate operands of casts. ... Cast operators in an
> integer constant expression shall only convert arithmetic types to
> integer types, ...", so (long)0.0 is permitted."
>
> While (void*) looks like a cast, in this context it is a compound
> literal of pointer type, which is not allowed.  [..]

Don't be silly.  The compound literal works just fine in the
context I mentioned for it:

    #include <stdio.h>

    int
    main(){
        printf( " null pointer : %p\n", (void*){ (long)0.0 } );
        return  0;
    }

Compile it for yourself if you don't believe me.